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Chapter 5: Problem 5

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.)

Short Answer

Expert verified
The angle \( \theta \) must be approximately \( 48.19^\circ \).

Step by step solution

01

Analyze the Forces

Consider the forces acting on the picture frame. There is the weight of the frame acting downward and the tension in the two wires, each pulling at an angle \( \theta \) from the vertical.
02

Set up the Equations

Let \( W \) be the weight of the frame. The vertical component of the tension in each wire will balance the weight of the picture. If \( T \) is the tension in one wire, and given that \( T = 0.75W \), the sum of the vertical components can be written as: \( 2T \cos \theta = W \).
03

Substitute Known Values

Substituting \( T = 0.75W \) into the equation: \( 2(0.75W) \cos \theta = W \). Simplify this to: \( 1.5W \cos \theta = W \).
04

Solve for \( \cos \theta \)

Divide both sides of the equation by \( W \): \( 1.5 \cos \theta = 1 \). Thus, \( \cos \theta = \frac{1}{1.5} = \frac{2}{3} \).
05

Calculate \( \theta \)

Find \( \theta \) by calculating the inverse cosine: \( \theta = \cos^{-1}(\frac{2}{3}) \).
06

Conclusion

Using a calculator, \( \theta \approx 48.19^\circ \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Forces in Equilibrium
In physics, the concept of forces in equilibrium refers to a situation where all acting forces on an object are balanced, resulting in no net force. This means the object remains at rest or continues to move with constant velocity. In the context of the picture frame scenario, the frame is suspended and stationary, implying it is in equilibrium. There are two key factors at play:
  • Weight: This is the gravitational force acting downward on the picture frame. It is a result of the mass of the frame multiplied by the acceleration due to gravity.
  • Tension: Each wire exerts an upward force to counterbalance the frame's weight, ensuring the net force is zero. Since the system is at rest, the combined upward forces of the wires' tensions equal the downward gravitational force.
This state of balance requires that all vertical components of the forces sum to zero. Therefore, understanding equilibrium helps in setting up the correct equations that describe the system.
Tension in Ropes
Tension is a crucial force when dealing with ropes or wires, especially in situations involving objects being suspended. In our problem, tension refers to the force exerted along the length of each wire that holds the picture frame. Key characteristics of tension include:
  • Uniform Tension: Each wire carries the same tension force due to symmetry and equal angles with the vertical.
  • Relationship with Weight: The given tension in each wire is 0.75 times the weight of the frame. This relationship is instrumental in calculating the required angle that the wires make with the vertical.
  • Direction: The tension force is directed along the rope or wire away from the object it supports, balancing components must be computed to analyze its effect.
Understanding tension is vital as it allows us to determine the necessary conditions to maintain equilibrium, ensuring the frame remains stationary while suspended.
Trigonometry in Physics
Trigonometry is an essential mathematical tool in physics, especially in analyzing problems involving angles and forces. It helps us break down complex vectors into their components, which is crucial for solving equilibrium problems.For the picture frame:
  • Components of Force: Each of the tensions in the wires can be divided into vertical and horizontal components using trigonometric functions. The vertical component contributes to balancing the frame's weight.
  • Cosine Function: In this problem, we use the cosine function to relate the vertical component of tension to the weight. Solving for \(\theta\) from \(\cos \theta = \frac{2}{3}\) illustrates this application.
  • Inverse Trigonometric Functions: Finding the angle \(\theta\) by calculating the inverse cosine allows us to determine the exact angle each wire makes with the vertical, which was around 48.19 degrees.
Thus, trigonometry provides the necessary methods to dissect force vectors, making it integral in physics problem-solving.

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Most popular questions from this chapter

A machine part consists of a thin 40.0 -cm-long bar with small 1.15 -kg masses fastened by screws to its ends. The screws can support a maximum force of 75.0 \(\mathrm{N}\) without pulling out. This bar rotates about an axis perpendicular to it at its center. (a) As the bar is turning at a constant rate on a horizontal, frictionless surface, what is the maximum speed the masses can have without pulling out the screws? (b) Suppose the machine is redesigned so that the bar turns at a constant rate in a vertical circle. Will one of the screws be more likely to pull out when the mass is at the top of the circle or at the bottom? Use a free-body diagram to see why. (c) Using the result of part (b), what is the greatest speed the masses can have without pulling a screw?

A \(2540-\mathrm{kg}\) test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by \(v(t)=\) \(A t+B t^{2},\) where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.50 \(\mathrm{m} / \mathrm{s}^{2}\) and 1.00 s later an upward velocity of 2.00 \(\mathrm{m} / \mathrm{s}\) (a) Determine \(A\) and \(B\) , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

Two objects with masses 5.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) hang 0.600 \(\mathrm{m}\) above the floor from the ends of a cord 6.00 \(\mathrm{m}\) long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00 -kg object.

You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end (Fig. E5.46). Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 \(\mathrm{m}\) from the central shaft. (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of \(30.0^{\circ}\) with the vertical. (b) Does the angle depend on the weight of the passenger for a given rate of revolution?
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